Question

我正在尝试编写一个程序，要求用户输入整数序列。输入数字时，程序应计算奇数和偶数。用户输入完数据后，程序应显示输入的奇数和偶数。

我可以在不使用哨兵值的情况下实现上述结果吗？另外，如果我问用户“您想输入多少个数字？”，如果输入为“ 5”，我如何让程序输出“输入数据：” 5次？

#include <iostream>

using namespace std;

int main()
{
int userInput;
int evenCount = 0;
int oddCount = 0;
cout << "Odd and Evens ";

do
{
    cout << "Please enter an integer: ";
    cin >> userInput;


    if (userInput % 2 == 0)
    {
        evenCount++;
    }
    else
    {
        oddCount++;
    }

} while (userInput != 0);
cout << "You entered " << oddCount << " odd numbers, and " << evenCount - 1 << " even numbers " << endl;

system("pause");
return 0;
}

这是我要通过程序实现的目标：

How many numbers are you going to enter? 5
Enter data: -5
Enter data: 2
Enter data: 3
Enter data: 6
Enter data: 0
Odd numbers entered: 2
Even numbers entered: 3

Answer 1

首先提示用户要输入多少个数字。输入该数字后，请循环使用，直到您的奇数和偶数总数小于该数字为止。

赞：

#include <iostream>

using namespace std;

int main()
{
    int numberOfInput;
    cout<<"How many numbers are you going to enter? ";
    cin>>numberOfInput;

    int userInput;
    int evenCount = 0;
    int oddCount = 0;

    do
    {
        cout << "Enter data: ";
        cin >> userInput;


        if (userInput % 2 == 0)
        {
            evenCount++;
        }
        else
        {
            oddCount++;
        }

    }
    while (evenCount+oddCount<numberOfInput);

    cout<<"Odd numbers entered: "<<oddCount<<endl;

    cout<<"Even numbers entered: "<<evenCount<<endl;
    return 0;
}

Answer 2

我将使用一个简单的for循环。

int main() {
 int userinput;
 int even=0;
 int odd=0;
 cout<<"Enter a value: ";
 cin>>userinput;
 int integerAdded;
 for (int i=0;i<userinput;i++){
    cout<<"Enter Data: ";
    cin>>integerAdded;
    if ((integerAdded%2)==0){
      even++;
    }
    else{
      odd++;
    }
  }

  cout<<"The amount of even numbers is "<<even<<endl;
  cout<<"The number of odd characters is: "<<odd<<endl;
}

如果用户想输入x数量的数字，我该如何跟踪它并结束程序？

2 个答案: